Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
// Template {{{
#include <bits/stdc++.h>
#define REP(i,n) for(int i=0; i<(int)(n); ++i)
using namespace std;
typedef long long LL;

#ifdef LOCAL
#include "contest.h"
#else
#define dump(x) 
#endif

class range {
    struct Iterator {
        int val, inc;
        int operator*() {return val;}
        bool operator!=(Iterator& rhs) {return val < rhs.val;}
        void operator++() {val += inc;}
    };
    Iterator i, n;
    public:
    range(int e) : i({0, 1}), n({e, 1}) {}
    range(int b, int e) : i({b, 1}), n({e, 1}) {}
    range(int b, int e, int inc) : i({b, inc}), n({e, inc}) {}
    Iterator& begin() {return i;}
    Iterator& end() {return n;}
};

const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
inline bool valid(int x, int w) { return 0 <= x && x < w; }

void iostream_init() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.setf(ios::fixed);
    cout.precision(12);
}
//}}}

struct Edge{ int dst, cost; };
typedef vector<Edge> Node;
typedef vector<Node> Graph;
const int INF = 1000000000;
void add_edge(Graph& G, int a, int b) {
    G[a].push_back({b, 1});
    G[b].push_back({a, 1});
}
vector<int> SPFA(const Graph& G, int s) {
    int n = G.size();
    vector<int> dist(n, INF);
    vector<bool> inque(n);
    queue<int> que;

    dist[s] = 0;
    que.push(s);
    inque[s] = true;
    while(!que.empty()){
        int v = que.front();
        que.pop();
        inque[v] = false;
        for(Edge e : G[v]) {
            if(dist[e.dst] > dist[v] + e.cost) {
                dist[e.dst] = dist[v] + e.cost;
                if(!inque[e.dst]) {
                    que.push(e.dst);
                    inque[e.dst] = true;
                }
            }
        }
    }
    return dist;
}
int main(){
    iostream_init();
    int N, M, s, t;
    while(cin >> N >> M >> s >> t) {
        s--; t--;
        Graph G(N);
        REP(i, M) {
            int a, b;
            cin >> a >> b;
            a--; b--;
            add_edge(G, a, b);
        }

        vector<int> ds = SPFA(G, s);
        vector<int> dt = SPFA(G, t);
        unordered_map<int, int> cnt;
        for(int x : dt) cnt[x]++;
        int D = ds[t];
        long long ans = 0;
        REP(i, N) {
            // ds[i] + 1 + E == D-1
            int E = D - ds[i] - 2;
            ans += cnt[E];
        }
        cout << ans << endl;

    }

    return 0;
}

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